Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define INF 1e6
using namespace std;
typedef long long ll;
vector<int> G[100001];
int n,m,s,t;
int dist_s[100001];
int dist_l[100001];
int main(void){
	scanf("%d %d %d %d",&n,&m,&s,&t);
	s--;
	t--;
	for(int i=0;i<m;i++){
		int a,b;
		scanf("%d %d",&a,&b);
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	fill(dist_s,dist_s+n,INF);
	queue<int> que;
	que.push(s);
	dist_s[s]=0;
	while(que.size()){
		int v=que.front();
		que.pop();
		for(int i=0;i<G[v].size();i++){
			int nec=G[v][i];
			if(dist_s[nec]==INF){
				dist_s[nec]=dist_s[v]+1;
				que.push(nec);
			}
		}
	}
	fill(dist_l,dist_l+n,INF);
	que.push(t);
	dist_l[t]=0;
	while(que.size()){
		int v=que.front();
		que.pop();
		for(int i=0;i<G[v].size();i++){
			int nec=G[v][i];
			if(dist_l[nec]==INF){
				dist_l[nec]=dist_l[v]+1;
				que.push(nec);
			}
		}
	}
	int mincost=dist_s[t];
	ll cnt=0;
	sort(dist_s,dist_s+n);
	sort(dist_l,dist_l+n);
	for(int i=0;i<n;i++){
		int midl=lower_bound(dist_l,dist_l+n,mincost-dist_s[i]-2)-dist_l;
		int midr=upper_bound(dist_l,dist_l+n,mincost-dist_s[i]-2)-dist_l;
		if(midl<n && midl>=0 && dist_s[i]+dist_l[midl]+2==mincost){
			cnt+=(ll)(midr-midl);
		}
	}
	printf("%lld\n",cnt);
	return 0;
}